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Question Number 131202 by abdurehime last updated on 02/Feb/21

1 lim_(n→∞) (((3^n +2^n )/6^n )) and ^( lim _(n→∞) (((1+2^2 +32+4^2 +.......+n^2 )/n^3 ))) help me

$$\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \left(\frac{\mathrm{3}^{\mathrm{n}} +\mathrm{2}^{\mathrm{n}} }{\mathrm{6}^{\mathrm{n}} }\right)\:\mathrm{and} \\ $$$$\:\:\:\overset{\:\mathrm{lim}\:\underset{\mathrm{n}\rightarrow\infty} {\:}\left(\frac{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{32}+\mathrm{4}^{\mathrm{2}} +.......+\mathrm{n}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} }\right)} {\:} \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$

Commented by liberty last updated on 02/Feb/21

lim_(n→∞) (((1^2 +2^2 +3^2 +4^2 +...+n^2 )/n^3 ))

$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +...+\mathrm{n}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} }\right)\: \\ $$

Answered by liberty last updated on 02/Feb/21

lim_(n→∞) ((6^n (((3/6))^n +((2/6))^n ))/6^n ) = lim_(n→∞) [((1/2))^n +((1/3))^n ] = 0 + 0 = 0

$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6}^{\mathrm{n}} \left(\left(\frac{\mathrm{3}}{\mathrm{6}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{2}}{\mathrm{6}}\right)^{\mathrm{n}} \right)}{\mathrm{6}^{\mathrm{n}} }\:=\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left[\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{n}} \right] \\ $$$$\:=\:\mathrm{0}\:+\:\mathrm{0}\:=\:\mathrm{0} \\ $$

Commented by EDWIN88 last updated on 03/Feb/21

it is right answer !

$${it}\:{is}\:{right}\:{answer}\:! \\ $$

Answered by Chhing last updated on 02/Feb/21

=lim_(x→∞) ((((1/6)n(n+1)(2n+1))/n^3 )) =(1/6)lim_(x→∞) (((n^3 (2+(3/n)+(1/n^2 )))/n^3 )) =(1/6)×lim_(x→∞) (2+(3/n)+(1/n^2 )) =(1/6)×2 =(1/3)

$$ \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\frac{\mathrm{1}}{\mathrm{6}}\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{n}^{\mathrm{3}} }\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{n}^{\mathrm{3}} \left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)}{\mathrm{n}^{\mathrm{3}} }\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}×\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{2} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Answered by Ar Brandon last updated on 02/Feb/21

1.L=lim_(n→∞) (((3^n +2^n )/6^n ))=lim_(n→∞) {((3/6))^n +((2/6))^n }=0 2.T=lim_(n→∞) {((1+2^2 +3^2 +4^2 +∙∙∙+n^2 )/n^3 )}=lim_(n→∞) Σ_(k=1) ^n (k^2 /n^3 ) =lim_(n→∞) {(1/n)Σ_(k=1) ^n (k^2 /n^2 )}=∫_0 ^1 x^2 dx=(1/3)

$$\mathrm{1}.\mathscr{L}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{3}^{\mathrm{n}} +\mathrm{2}^{\mathrm{n}} }{\mathrm{6}^{\mathrm{n}} }\right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\left(\frac{\mathrm{3}}{\mathrm{6}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{2}}{\mathrm{6}}\right)^{\mathrm{n}} \right\}=\mathrm{0} \\ $$$$\mathrm{2}.\mathscr{T}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{n}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} }\right\}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right\}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2}} \mathrm{dx}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$

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